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Math 5120 Homework 2.1 Solutions 1. If D = f (c), and we also suppose that J = −D∇c (i.e. Fick’s law holds) and σ = 0 (i.e. no sources/sinks), then the 1d diffusion equation is ∂ ∂c ∂c =− −D ∂t ∂x ∂x ∂c ∂ −f (c) =− ∂x ∂x ∂c ∂c ∂2c = f ′ (c) · · + f (c) 2 ∂x ∂x ∂x 2 2 ∂c ∂ c . = D 2 + f ′ (c) ∂x ∂x 2. To derive the two-dimensional conservation equation, first let the flux be represented by J(x, y) = (Jx , Jy ). The number of particles in time ∆t that cross through a given area is defined by N = (J · m)A∆t, where m is normal to the given area. Therefore, we can think of the concentration as the number of particles per volume. If we consider a small rectangular region, where particles can only move in the x- and y-directions, we can define the volume of the region to be ∆V = ∆x·∆y·h, where h is the height of the region. Then the change in the concentration, ∆c, in the small rectangular region beginning at (x0 , y0 ) is given by ∆N [J(x0 , y0 ) · (1, 0)]∆y · h · ∆t [J(x0 + ∆x, y0 ) · (1, 0)]∆y · h · ∆t = − ∆V ∆x · ∆y · h ∆x · ∆y · h [J(x0 , y0 ) · (0, 1)]∆x · h · ∆t [J(x0 , y0 + ∆y) · (0, 1)]∆x · h · ∆t − + ∆x · ∆y · h ∆x · ∆y · h Jx (x0 , y0 ) − Jx (x0 + ∆x, y0 ) Jy (x0 , y0 ) − Jy (x0 , y0 + ∆y) ∆c = + ± σ(x, y) ⇒ ∆t ∆x ∆y ∂c ∂Jx ∂Jy ⇒ =− + ± σ(x, y) = −∇ · J ± σ ∂t ∂x ∂y ∆c = 3. (a) The initial flux is given by J = c0 α∇Ψ, which, if α = 1 and c0 = 4, is J0 = (−4x, −4y). (b) If a bar of unit length is placed at (1, 1) parallel to the x-axis, then the normal vector for particles being attracted to the origin is (0, −1). Therefore, the number of particles crossing this bar during a time interval of length ∆t is N = [J0 · (0, −1)]1∆t = 4∆t, where it’s assumed that ∆t is a small interval just after the field is turned on. 1 Homework 2.2 Solutions L2 2D ln La be associated with random motion and T = Ls be associated with direct q motion, where L = A π . Recall that the bacterial doubling time is about 5 h. 1. (a) Let τ = i. If the macrophages move twice as fast, then s = 6, and τ ≈ 20 h and T ≈ 47 min. ii. If the target (surface area) is twice as big, then a = 40, and τ ≈ 58 h and T ≈ 1.5 h. iii. If the area of the alveolus is half as big, then A = 0.5(2.5 × 105 ), and τ ≈ 34 h and T ≈ 1 h. In all cases, direct motion toward the target occurs faster than the bacteria double, while random motion is a considerably slower process. (b) L2 L s τ = ln · T 2D a√ L √ L L A = ln = − √ ln( Aπ) ǫs a ǫs π R0 = √ √ √ If R0 = 1, then s = − ǫ√Aπ ln( Aπ). If s > 0, then it must be the case that ln( Aπ) < 0, which can only happen provided A < π1 . So, in order for random motion to be as fast as direct motion, the area in question must be sufficiently small. 2. The steady-state solution of the 1d diffusion equation satisfies ∂2c d2 c ∂c =D 2 =0 ⇒ = 0. ∂t ∂x dx2 This differential equation has solution c(x) (no t dependence) subject to the given boundary conditions. To solve for c(x), we integrate: dc d2 c =0 ⇒ = k1 ⇒ c(x) = k1 x + k2 . 2 dx dx To solve for constants k1 and k2 , use the boundary conditions: c(0) = k2 = 0 and c(L) = k1 L = C0 ⇒ k1 = CL0 . Therefore, x c(x) = k1 x + k2 = C0 , L C0 which is a line with slope L that passes through the origin. c(x) C0 0 L 0 x The total number of particles in the interval [0, L] is Z L Z L x C0 1 2 L C0 1 2 1 C0 dx = c(x) dx = x · L = C0 L. = L L 2 L 2 2 0 0 0 2